Reciprocity Theorems for Bettin--Conrey Sums

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by  Juan S. Auli

RECIPROCITY THEOREMS FOR BETTIN–CONREY SUMS
6 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK
1
0 DedicatedtothememoryofTomM.Apostol
2

g ABSTRACT. RecentworkofBettinandConreyontheperiodfunctionsofEisensteinseriesnaturallygaverise
u totheDedekind-likesum
A ca h = kak(cid:229)−1cot p mh z −a,m ,
k k k
3 (cid:18) (cid:19) m=1 (cid:18) (cid:19) (cid:16) (cid:17)
wherea∈C,handkarepositivecoprimeintegers,andz (a,x)denotestheHurwitzzetafunction. Wederive

] anewreciprocitytheoremfortheseBettin–Conreysums,whichinthecaseofanoddnegativeintegeracanbe
T
explicitlygivenintermsofBernoullinumbers. This,inturn,impliesexplicitformulasfortheperiodfunctions
N appearinginBettin–Conrey’swork.WestudygeneralizationsofBettin–Conreysumsinvolvingzetaderivatives
. andmultiplecotangentfactorsandrelatethesetospecialvaluesoftheEstermannzetafunction.
h
t
a
m
[ 1. INTRODUCTION AND STATEMENT OF RESULTS

2 OurpointofdepartureisrecentworkofBettinandConrey[7,8]ontheperiodfunctionsofEisensteinse-
v
ries. TheirinitialmotivationwasthederivationofanexactformulaforthesecondmomentsoftheRiemann
9
3 zetafunction, buttheirworknaturally gaverisetoafamilyoffinitearithmeticsumsoftheform
8
6 c h = kak(cid:229)−1cot p mh z −a,m ,
0 a k k k
1. (cid:18) (cid:19) m=1 (cid:18) (cid:19) (cid:16) (cid:17)
wherea∈C,handkarepositivecoprimeintegers, andz (a,x)denotes theHurwitzzetafunction
0
16 z (a,x) = (cid:229)¥ 1 ,
: (n+x)a
v n=0
i initially defined for ´ (a)>1and meromorphically continued to the a-plane. Wecall c (h)and its natural
X a k
generalizations appearing belowBettin–Conrey sums.
r
a Therearetwomajormotivationstostudythesesums. Thefirstisthatc (h)isessentiallyaVasyuninsum,
0 k
which in turn makes a critical appearance in the Nyman–Beurling–Ba´ez-Duarte approach to the Riemann
hypothesis through the twisted mean-square of the Riemann zeta function on the critical line (see, e.g.,
[5,16]). Bettin–Conrey’s work,fora=0,impliesthatthereisahiddensymmetryofthismean-square.
The second motivation, and the central theme of our paper, is that the Bettin–Conrey sums satisfy a
reciprocity theorem:
h k 1+a −k kaaz (1−a)
c − c +
a k h a h p h
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
Date:2August2016.
2010MathematicsSubjectClassification. Primary11F20;Secondary11L03,11M35.
Keywordsandphrases. Dedekindsum,cotangentsum,Bettin–Conreysum,reciprocitytheorem,Hurwitzzetafunction,period
function,quantummodularform,Estermannzetafunction.
We thank Sandro Bettin for valuable comments, particularly for suggesting Theorem 3. Abdelmejid Bayad was partially
supportedbytheFDIRoftheUniversite´d’EvryVald’Essonne;MatthiasBeckwaspartiallysupportedbytheUSNationalScience
Foundation(DMS-1162638).
1
2 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK
extends from its initiation domain Q to an (explicit) analytic function on C\R , making c nearly an
≤0 a
example of a quantum modular form in the sense of Zagier [18]. In fact, Zagier’s “Example 0” is the
Dedekindsum
1 k(cid:229)−1 p mh p m
s(h,k) = cot cot ,
4k k k
m=1 (cid:18) (cid:19)
(cid:16) (cid:17)
whichis,uptoatrivialfactor, c (h). Dedekind sumsfirstappeared inthetransformation properties ofthe
−1 k
Dedekindetafunction andsatisfythereciprocity theorem [10,11]
1 1 h 1 k
s(h,k)+s(k,h) = − + + + .
4 12 k hk h
(cid:18) (cid:19)
WenowrecallthepreciseformofBettin–Conrey’s reciprocity theorem.
Theorem1(Bettin–Conrey [7]). Ifhandkarepositivecoprimeintegers then
h k 1+a −k kaaz (1−a) h
c − c + = −iz (−a)y
a k h a h p h a k
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
where
i z (1−a) i p a g (z)
y (z) = − cot +i a
a p z z (−a) z1+a 2 z (−a)
and
g (z) = −2 (cid:229) (−1)n B2n z (1−2n−a)(2p z)2n−1
a
(2n)!
1≤n≤M
1 cosp a
+ z (s)z (s−a)G (s) 2 (2p z)−sds.
p iZ(−21−2M) sinp s−2a
Here B denotes the kth Bernoulli number, M is any integer ≥−1min(0,´ (a)), and the integral notation
k 2
indicates thatourintegration pathisovertheverticalline´ (s)=−1−2M.
2
WenotethatBettinandConreyinitiallydefinedy (z)through
a
1 1
y (z) = E (z)− E − ,
a a+1 za+1 a+1 z
(cid:18) (cid:19)
inotherwords,y (z)istheperiodfunction oftheEisenstein seriesofweight2k+2,
a
E (z) = 1+ 2 (cid:229) s (n)e2p inz,
a+1 z (−a) a
n≥1
wheres (n)=(cid:229) da,andthenshowedthaty (z)satisfiestheproperties ofTheorem1.
a d|n a
We have several goals. We start by showing that the right-hand side of Theorem 1 can be simplified by
employinganintegration techniqueforDedekind-like sumsthatgoesbacktoRademacher[11]. Thisyields
ourfirstmainresult:
Theorem 2. Let ´ (a) > 1 and suppose h and k are positive coprime integers. Then for any 0 < e <
min 1,1 ,
h k
(cid:8) (cid:9) h k az (a+1) (hk)1−a cot(p hz)cot(p kz)
h1−ac +k1−ac = − dz.
−a k −a h p (hk)a 2i (e) za
(cid:18) (cid:19) (cid:18) (cid:19) Z
Theorem 2 implies the following analytic continuation result; in particular, in this sense Theorem 2 can
beextendedtoallcomplexa.
RECIPROCITYTHEOREMSFORBETTIN–CONREYSUMS 3
Theorem3. Lethandkarepositivecoprimeintegers. Thenforany0<e <min 1,1 ,thefunction
h k
cot(p hz)cot(p kz) (cid:8) (cid:9)
F(a) = dz
(e) za
Z
hasaholomorphic continuation tothewholecomplexplane.
Second, we employ Theorem 2 to show that in the case that a is an odd negative integer, the right-hand
sideofthereciprocity theorem canbeexplicitly givenintermsofBernoullinumbers.
Theorem4. Letn>1beanoddinteger andsuppose handkarepositive coprimeintegers. Then
h1−nc h +k1−nc k = 2p i n 1 nB +n(cid:229)+1 n+1 B B hmkn+1−m .
−n −n n+1 m n+1−m
k h hk i(n+1)! m
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) m=0(cid:18) (cid:19) !
Wenotethat,asfaraswecantell,thisidentitycannotbeeasilyderivedfromTheorem1. Ourthirdmain
result is, in turn, a consequence of Theorem 4: in conjunction with Theorem 1, it implies the following
explicitformulasfory (z)andg (z)whenaisanoddnegativeinteger.
a a
Theorem5. Ifn>1isanoddinteger thenforallz∈C\R
≤0
y (z) = (2p i)n n(cid:229)+1 n+1 B B zm−1
−n z (n)(n+1)! m m n+1−m
m=0(cid:18) (cid:19)
and
g (z) = (2p i)n (cid:229) n n+1 B B zm.
−n m+1 n−m
i(n+1)! m+1
m=0(cid:18) (cid:19)
In [7, Theorem 2], Bettin and Conrey computed the Taylor series of g (z) and remarked that, if a is a
a
negativeinteger,p g(m)(1)isarationalpolynomialinp 2. Theorem5generalizesthisremark. Wewillprove
a
Theorems2–5inSection2.
Our next goal is to study natural generalizations of c (h). Taking a leaf from Zagier’s generalization of
a k
s(h,k) to higher-dimensional Dedekind sums [17] and its variation involving cotangent derivatives [9], let
k ,k ,...,k be positive integers such that (k ,k )=1 for j=1,2,...,n, let m ,m ,...,m be nonnegative
0 1 n 0 j 0 1 n
integers, a6=−1acomplexnumber,anddefinethegeneralized Bettin–Conrey sum
c k0 k1 ··· kn = kak(cid:229)0−1z (m0) −a, l (cid:213) n cot(mj) p kjl .
a(cid:18) m0 m1 ··· mn (cid:19) 0 l=1 (cid:18) k0(cid:19)j=1 (cid:18) k0 (cid:19)
Herez (m0)(a,z)denotesthemth derivativeoftheHurwitzzetafunction withrespect toz.
0
Thisnotation mimicsthatofDedekindcotangent sums;notethat
h k h
c = c .
s k s 0 0
(cid:18) (cid:19) (cid:18) (cid:19)
In Section 3, we will prove reciprocity theorems for generalized Bettin–Conrey sums, paralleling Theo-
rems2and4,aswellasmorespecialcasesthatgive,wethink, interesting identities.
Ourfinalgoalistorelatetheparticular generalized Bettin–Conrey sum
q−1
(cid:229) cot(k)(p mx)z −a,m
q
m=1
(cid:16) (cid:17)
withevaluations oftheEstermannzetafunction (cid:229) s (n)e2pinx atintegers s;seeSection4.
n≥1 a ns
4 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK
2. PROOFS OF MAIN RESULTS
InordertoproveTheorem2,weneedtwolemmas.
Lemma6. Letmbeanonnegative integer. Then
∓i ifm=0,
limcot(m)p (x±iy) =
y→¥ (0 ifm>0.
Furthermore, thisconvergence isuniform withrespecttoxinafixedboundedinterval.
Proof. Sincecotz= i(eiz+e−iz),wemayestimate
eiz−e−iz
2 2 2
|i+cotp (x+iy)| = ≤ = .
ei(2p x)−e2p y ei(2p x) −|e2p y| |1−e2p y|
Giventhattherightmostterminthisin(cid:12)equality vanis(cid:12)hesa(cid:12)s(cid:12)y→¥ (cid:12),wesee(cid:12)that
(cid:12) (cid:12) (cid:12)(cid:12) (cid:12) (cid:12)
limcotp (x+iy) = −i.
y→¥
Similarly,theinequality
2 2 2
|−i+cotp (x−iy)| = ≤ =
ei(2p x)e2p y−1 ei(2p x)e2p y −1 |e2p y−1|
impliesthatlimy→¥ cotp (x−iy)=i. Si(cid:12)nce (cid:12) (cid:12)(cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12)(cid:12) (cid:12) (cid:12)
2ep y 2ep y 2ep y
|cscp (x+iy)| = ≤ = ,
|eip x−e−ip xe2p y| ||eip x|−|e−ip xe2p y|| |1−e2p y|
itfollowsthatlimy→¥ cscp (x+iy)=0. Similarly,
2ep y 2ep y 2ep y
|cscp (x−iy)| = ≤ =
|eip xe2p y−e−ip x| ||eip xe2p y|−|e−ip x|| |e2p y−1|
impliesthatlimy→¥ cscp (x−iy)=0. Weremarkthat ddz(cotz)=−csc2zand
d
(cscz) = −csczcotz,
dz
soallthederivativesofcotzhaveacsczfactor, andtherefore,
∓i ifm=0,
limcot(m)p (x±iy) =
y→¥ (0 ifm>0.
Since the convergence above is independent of x, the limit is uniform with respect to x in a fixed bounded
interval. (cid:3)
Lemma6impliesthat
∓i ifm=0,
limcot(m)p h(x±iy) = limcot(m)p k(x±iy) =
y→¥ y→¥ (0 ifm>0,
uniformlywithrespecttoxinafixedbounded interval.
TheproofofthefollowinglemmaishintedatbyApostol[3].
Lemma7. If´ (a)>1andR>0,thenz (a,x+iy)vanishesuniformlywithrespecttox∈[0,R]asy→±¥ .
RECIPROCITYTHEOREMSFORBETTIN–CONREYSUMS 5
Proof. Webeginbyshowingthatz (a,z)vanishesas` (z)→±¥ if´ (z)>0. Since´ (a)>1and´ (z)>0,
wehavetheintegral representation [14,eq.25.11.25]
1 ¥ ta−1e−zt
z (a,z) = dt,
G (a) 1−e−t
0
Z
whichmaybewrittenas
1 ¥ ta−1e−t´ (z)
(1) z (a,z) = e−it` (z)dt.
G (a) 1−e−t
0
Z
Notethatforfixed´ (z),
¥ ta−1e−t´ (z)
dt = z (a,´ (z))G (a)
1−e−t
0
Z
and
¥ ta−1e−t´ (z) ¥ t´ (a)−1e−t´ (z)
dt = dt = z (´ (a),´ (z))G (´ (a)),
1−e−t 1−e−t
0 (cid:12) (cid:12) 0
Z Z
(cid:12) (cid:12)
sotheRiemann–Lebe(cid:12)sgue lemma(cid:12)(see,forexample,[13,Theorem16])impliesthat
(cid:12) (cid:12)
(cid:12) (cid:12) ¥ ta−1e−t´ (z)
e−it` (z)dt
1−e−t
0
Z
vanishes as ` (z)→±¥ . By (1), this means that for ´ (z) fixed, z (a,z) vanishes as ` (z)→±¥ . In other
words,z (a,x+iy)→0pointwisewithrespect tox>0asy→±¥ .
Moreover, the vanishing of z (a,x+iy) as y→±¥ is uniform with respect to x∈[0,R]. Indeed, denote
g(t)= ta−1e−tR,then(1)impliesthat
1−e−t
¥
g(t)dt = G (a)z (a,R)
0
Z
and
¥
|g(t)|dt = G (´ (a))z (´ (a),R).
0
It then follows from the Riemann–ZLebesgue lemma that lim|z|→¥ 0¥ g(t)e−itzdt =0. If x∈(0,R], we may
write
¥ ta−1e−tx ¥R
G (a)z (a,x±iy) = e∓itydt = g(t)e−it(±y−i(x−R))dt.
1−e−t
0 0
Z Z
Sinceg(t)doesnotdependonx,thespeedatwhichz (a,x±iy)vanishesdependsonRandy2+(x−R)2.
However,weknowthat0≤|x−R|<R,sothespeedofthevanishing dependsonlyonR.
Finally,notethat
¥ ¥
z (a,iy) = (cid:229) 1 = (cid:229) 1 + 1 = z (a,1+iy)+ 1 ,
(iy+n)a (1+iy+n)a (iy)a (iy)a
n=0 n=0
so z (a,iy)→0 as y→±¥ , and the speed at which z (s,iy) vanishes depends on that of z (s,1+iy). Thus,
z (s,x+iy)→0uniformly asy→±¥ ,aslongasx∈[0,R]. (cid:3)
ProofofTheorem2. TheideaistouseCauchy’sresiduetheorem tointegrate thefunction
f(z) = cot(p hz)cot(p kz)z (a,z)
alongC(M,e )asM→¥ ,whereC(M,e )denotes thepositively oriented rectangle withvertices 1+e +iM,
e +iM,e −iM and1+e −iM,forM>0and0<e <min 1,1 (seeFigure1).
h k
Henceforth, a ∈ C is such that ´ (a) > 1, (h,k) is a pair of coprime positive intergers, and f(z) and
C(M,e )areasabove,unlessotherwisestated. Sincez (a,z)(cid:8)isana(cid:9)lyticinsideC(M,e ),theonlypolesof f(z)
6 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK
FIGURE 1. TheclosedcontourC(M,e ).
arethose ofthecotangent factors. Thus,thefactthathandk arecoprimeimpliesthatacompletelistofthe
possible polesof f(z)insideC(M,e )is
1 h−1 1 k−1
E = ,..., , ,..., ,1
h h k k
(cid:26) (cid:27)
and each of these poles is (at most) simple, with the exception of 1, which is (at most) double. For m∈
{1,2,...,h−1},
p km m 1 1 p km m
Resf(z) = cot cos(p m)z a, Res = cot z a, .
z=m h h z=m sin(p hz) p h h h
h (cid:18) (cid:19) h (cid:18) (cid:19)
(cid:16) (cid:17) (cid:16) (cid:17)
Ofcourse,ananalogous resultistrueforResz=m f(z)forallm∈{1,2,...,k−1},andtherefore
k
(cid:229) Resf(z) = Resf(z)+ 1 h(cid:229)−1cot p km z a,m + 1 k(cid:229)−1cot p hm z a,m
z0∈Ez=z0 z=1 p hm=1 (cid:18) h (cid:19) (cid:16) h(cid:17) p km=1 (cid:18) k (cid:19) (cid:16) k(cid:17)
or,equivalently,
(2) h1−ac h +k1−ac k = p (hk)1−a (cid:229) Resf(z) −Resf(z) .
−a −a
(cid:18)k(cid:19) (cid:18)h(cid:19) z0∈Ez=z0 ! z=1 !
RECIPROCITYTHEOREMSFORBETTIN–CONREYSUMS 7
Wenowdetermine Res f(z). TheLaurentseriesofthecotangent function about0isgivenby
z=1
1 1 1 2
cotz = − z− z3− z5+···,
z 3 45 945
so,bytheperiodicity ofcotz,forz6=1inasmallneighborhood ofz=1,
1 1 p k (p k)3 2(p k)5
cot(p kz) = − (z−1)− (z−1)3− (z−1)5+···
p k z−1 3 45 945
(cid:18) (cid:19)
and,similarly,
1 1 p h (p h)3 2(p h)5
cot(p hz) = − (z−1)− (z−1)3− (z−1)5+···.
p h z−1 3 45 945
(cid:18) (cid:19)
Sincez (a,z)isanalytic inasmallneighborhood of1,Taylor’stheoremimpliesthat
¥
z (a,z)= (cid:229) b (z−1)n,
n
n=0
where b = z (n)(a,1) for n=0,1,2,... (derivatives relative to z). Thus, the expansion of f(z) about 1 is of
n n!
theform
2
b 1 b 1
0 1
+ +(analyticpart).
p 2hk z−1 p 2hk z−1
(cid:18) (cid:19) (cid:18) (cid:19)
Giventhata6=0,1,weknowthat ¶ z (a,z)=−az (a+1,z)[14,eq.25.11.17], sob =−az (a+1,1)=
¶ z 1
−az (a+1). Weconclude thatRes f(z)=−az (a+1) anditthenfollowsfrom(2)that
z=1 p 2hk
(3) h1−ac h +k1−ac k = az (a+1)+ p (cid:229) Resf(z).
−a(cid:18)k(cid:19) −a(cid:18)h(cid:19) p (hk)a (hk)a−1z0∈Ez=z0
(cid:229)
We now turn to the computation of Resf(z) via Cauchy’s residue theorem, which together with (3)
z0∈Ez=z0
willprovide thereciprocity weare after. Notethat thefunction f(z) isanalytic onanytwoclosed contours
C(M ,e )andC(M ,e )and since the poles inside these twocontours arethe same, wemayapply Cauchy’s
1 2
residuetheorem tobothcontours anddeducethat
f(z)dz = f(z)dz.
ZC(M1,e) ZC(M2,e)
Inparticular, thisimpliesthat
(4) lim f(z)dz = 2p i (cid:229) Resf(z).
M→¥ ZC(M,e) z0∈Ez=z0
Letg bethepathalongC(M,e )from1+e +iMtoe +iM. Similarly,defineg frome −iMto1+e −iM,
1 2
g frome +iM toe −iM,andg from1+e −iM to1+e +iM (seeFigure1). Since´ (a)>1,
3 4
¥ ¥
z (a,z+1) = (cid:229) 1 = (cid:229) 1 = z (a,z)− 1 ,
(n+z+1)a (n+z)a za
n=0 n=1
andsotheperiodicity ofcotzimpliesthat
cot(p hz)cot(p kz)
f(z)dz = − f(z)dz+ dz.
g g g za
Z 4 Z 3 Z 3
8 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK
Lemmas6and7implythat f(z)vanishesuniformlyasM→¥ (uniformitywithrespectto´ (z)∈[e ,1+e ])
so
lim f(z)dz = 0 = lim f(z)dz.
M→¥ g M→¥ g
Z 1 Z 2
Thismeansthat
lim f(z)dz = lim f(z)dz+ f(z)dz
M→¥ ZC(M,e) M→¥ (cid:18)Zg3 Zg4 (cid:19)
anditfollowsfrom(3)and(4)that
h k az (a+1) (hk)1−a e−i¥ cot(p hz)cot(p kz)
h1−ac +k1−ac = + dz.
−a k −a h p (hk)a 2i e+i¥ za
(cid:18) (cid:19) (cid:18) (cid:19) Z
ThiscompletestheproofofTheorem2. (cid:3)
ProofofTheorem3. FromTheorem2,for´ (a)>1,
(hk)1−a az (a+1) h k
F(a) = − h1−ac +k1−ac ,
2i p (hk)a −a k −a h
(cid:18) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19)
sonowwestudythefunction a7→ az (a+1)− h1−ac h +k1−ac k .
p (hk)a −a k −a h
From the definition of c (h) (resp., c ((cid:0)k)) the fu(cid:0)nc(cid:1)tion h1−ac (cid:0) (cid:1)h(cid:1) +k1−ac k is a linear com-
−a k −a h −a k −a h
bination (with holomorphic coefficients) of the Hurwitz function z (s,x) with s = −a and x = m with
m=1,...,k−1(resp., s=−aandx= m withm=1,...,k−1). Thisg(cid:0)ive(cid:1)saholomor(cid:0)ph(cid:1)ic continuaktion to
h
C\{1}. However,ata=1,
h k
h1−ac +k1−ac = 2p (s(h,k)+s(k,h))
−a −a
k h
(cid:18) (cid:19) (cid:18) (cid:19)
iswelldefined. Thereforethefunction h1−ac h +k1−ac k hasaholomorphic continuation toC.
−a k −a h
Next,itisclearthatthefunction
s(cid:0)7→(cid:1)sz (s+1) (cid:0) (cid:1)
hasaholomorphiccontinuation toC\{0}. Theanalyticityofthisfunctionats=0followsfromtheanalytic
continuation oftheRiemannz -function.
Thus,ourdesiredresultfollowsfromthemonodromytheorem (see,e.g.,[15]). (cid:3)
Toprove Theorem 4, we now turn to the particular case in which a=n>1 is an odd integer and study
Bettin–Conrey sumsoftheformc .
−n
LetY (n)(z)denotethe(n+2)-thpolygammafunction(see,forexample,[14,Sec.5.15]). Itiswellknown
thatfornapositiveinteger,
(−1)n+1Y (n)(z)
z (n+1,z) =
n!
whenever´ (z)>0(see,forinstance, [14,eq.25.11.12]), soforn>1,wemaywrite
c h = (−1)n k(cid:229)−1cot p mh Y (n−1) m .
−n k kn(n−1)! k k
(cid:18) (cid:19) m=1 (cid:18) (cid:19)
(cid:16) (cid:17)
Bythereflectionformulaforthepolygammafunctions [14,eq.5.15.6],
Y (n)(1−z)+(−1)n+1Y (n)(z) = (−1)np cot(n)(p z),
weknowthatifnisodd,then
m m p m
Y (n−1) 1− −Y (n−1) = p cot(n−1)
k k k
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
RECIPROCITYTHEOREMSFORBETTIN–CONREYSUMS 9
foreachm∈{1,2,...,k−1}. Therefore,
2k(cid:229)−1cot p mh Y (n−1) m = k(cid:229)−1cot p mh Y (n−1) m
k k k k
m=1 (cid:18) (cid:19) m=1 (cid:18) (cid:19)
(cid:16) (cid:17) (cid:16) (cid:17)
+k(cid:229)−1cot p (k−m)h Y (n−1) 1−m ,
k k
m=1 (cid:18) (cid:19)
(cid:16) (cid:17)
whichimpliesthat
2k(cid:229)−1cot p mh Y (n−1) m = k(cid:229)−1cot p mh Y (n−1) m −Y (n−1) 1−m
k k k k k
m=1 (cid:18) (cid:19) m=1 (cid:18) (cid:19)
(cid:16) (cid:17) (cid:16) (cid:16) (cid:17) (cid:16) (cid:17)(cid:17)
= −p k(cid:229)−1cot p mh cot(n−1) p m .
k k
m=1 (cid:18) (cid:19)
(cid:16) (cid:17)
This means that for n> 1 odd, c is essentially a Dedekind cotangent sum. Indeed, using the notation
−n
in[9],
c h = p k(cid:229)−1cot p mh cot(n−1) p m
−n k 2kn(n−1)! k k
(cid:18) (cid:19) m=1 (cid:18) (cid:19)
(cid:16) (cid:17)
k h 1
p
= c n−1 0 n−1 .
2(n−1)!  
0 0 0
 
ThusTheorem 4isaninstance ofTheorem2. Itssignificance isareciprocity instance forBettin–Conrey
sumsoftheformc intermsofBernoullinumbers. Forthisreasonwegivethedetailsofitsproof.
−n
ProofofTheorem4. We consider the closed contour C(M,e ) defined as the positively oriented rectangle
with vertices 1+iM, iM, −iM and 1−iM, with indentations (to the right) of radius 0<e <min 1,1
h k
around0and1(seeFigure2). e
(cid:8) (cid:9)
SinceC(M,e )contains thesamepoles of f(z)=cot(p hz)cot(p kz)z (n,z) astheclosed contourC(M,e )
inFigure1usedtoproveTheorem2,wemayapplyCauchy’sresiduetheorem,lettingM→¥ ,andweonly
needtodeeterminelimM→¥ C(M,e) f(z)dzinordertodeduceareciprocity lawforthesumsc−n.
Asin the case ofC(M,e ), the integrals along the horizontal paths vanish, so using the periodicity of the
R
cotangent toaddintegralsaloengparallelpaths, aswedidwhenconsideringC(M,e ),weobtain
ie −iM
(5) lim f(z)dz = lim g(z)dz+ g(z)dz + g(z)dz,
M→¥ ZC(M,e) M→¥ (cid:18)ZiM Z−ie (cid:19) Zg3
whereg denotestheindented patharound 0and
3 e
cot(p kz)cot(p hz)
g(z) = .
zn
Giventhatg(z)isanoddfunction,theverticalintegralscancelandwemayapplyCauchy’sresiduetheorem
tointegrate g(z)alongthepositivelyoriented circleofradiuse andcentered at0,todeduce that
lim f(z)dz = −p iResg(z).
M→¥ C(M,e) z=0
Z
This is the main reason to use the contouer C(M,e ) instead ofC(M,e ). Indeed, integration along C(M,e )
exploitstheparityofthefunction g(z),allowingustocanceltheverticalintegrals in(5).
e e
10 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK
FIGURE 2. TheclosedcontourC(M,e ).
Theexpansion ofthecotangent function is e
p zcot(p z) = (cid:229)¥ (2p i)mBmzm,
m!
m=0
withtheconvention thatB mustberedefinedtobezero. Thus,wehavetheexpansion
1
cot(p kz) = (cid:229)¥ (2i)(2p ik)mBm+1zm
(m+1)!
m=−1
andofcourse, ananalogous resultholdsforh. Hence,
(6) Resg(z) = (2i)(2p i)n n(cid:229)+1 n+1 B B hmkn+1−m,
z=0 p hk(n+1)! m m n+1−m
m=0(cid:18) (cid:19)
andgiventhatz (n+1)=−(2p i)n+1B [14,eq.25.6.2],theCauchyresidue theoremand(3)yield
2(n+1)! n+1
(7) 2p i n 1 nB +n(cid:229)+1 n+1 B B hmkn+1−m = h1−nc h +k1−nc k .
n+1 m n+1−m −n −n
hk i(n+1)! m k h
(cid:18) (cid:19) m=0(cid:18) (cid:19) ! (cid:18) (cid:19) (cid:18) (cid:19)
Finally, notethat theconvention B :=0isirrelevant in(7),since B inthis sum isalways multiplied by
1 1
aBernoullinumberwithoddindexlargerthan1. (cid:3)
NotethatTheorem4isessentiallythesameasthereciprocitydeducedbyApostolforDedekind–Apostol
sums [2]. This is a consequence of the fact that for n > 1 an odd integer, c h is a multiple of the
−n k
Dedekind–Apostol sums (h,k). Indeed,forsuchn[3,Theorem1]
n
(cid:0) (cid:1)
h
s (h,k) = in!(2p i)−nc .
n −n
k
(cid:18) (cid:19)